Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For each of the following pairs, predict which substance is more soluble in water. a. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) or \(\mathrm{NH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{CN}\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) d. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) f. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) or \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Short Answer

Expert verified
a. NH₃ is more soluble. b. CH₃OCH₃ is more soluble. c. CH₃CH₂OH is more soluble. d. CH₃OH is more soluble. e. (CH₃)₃CCH₂OH is more soluble. f. CH₃CO₂H is more soluble.

Step by step solution

01

Identify polar functional groups

Both molecules possess a nitrogen atom and have the capability to form hydrogen bonds.
02

Compare hydrogen bonding capabilities

\(\mathrm{CH}_{3} \mathrm{NH}_{2}\) has the additional \(\mathrm{CH}_{3}\) group which makes it less polar than \(\mathrm{NH}_{3}\). Therefore, \(\mathrm{NH}_{3}\) is more soluble in water. b. Comparing \(\mathrm{CH}_{3} \mathrm{CN}\) and \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\):
03

Identify polar functional groups

Both molecules possess polar functional groups: \(\mathrm{CH}_{3} \mathrm{CN}\) has a nitrile group and \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) has an ether group.
04

Compare hydrogen bonding capabilities

\(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) has an oxygen atom that allows for hydrogen bonding, while \(\mathrm{CH}_{3} \mathrm{CN}\) does not. Therefore, \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) is more soluble in water. c. Comparing \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (ethanol) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (propane):
05

Identify polar functional groups

Ethanol has a polar hydroxyl group while propane has no polar functional groups.
06

Compare hydrogen bonding capabilities

Ethanol can form hydrogen bonds with water, while propane cannot. Therefore, ethanol is more soluble in water. d. Comparing \(\mathrm{CH}_{3} \mathrm{OH}\) (methanol) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (ethanol):
07

Identify polar functional groups

Both molecules possess polar hydroxyl groups and have the capability to form hydrogen bonds.
08

Compare size and polarity

Methanol is smaller and less "nonpolar" compared to ethanol. Therefore, methanol is more soluble in water. e. Comparing \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) and \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\):
09

Identify polar functional groups

Both molecules possess polar hydroxyl groups and have the capability to form hydrogen bonds.
10

Compare size and polarity

The second molecule has a longer nonpolar hydrocarbon chain, which reduces its solubility in water. Therefore, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) is more soluble in water. f. Comparing \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\):
11

Identify polar functional groups

\(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) has a polar ether group, and \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) has a polar carboxylic acid group.
12

Compare hydrogen bonding capabilities

\(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) has a stronger hydrogen bonding capability due to its carboxylic acid group, while \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) has a weaker hydrogen bonding capability. Therefore, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) is more soluble in water.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Functional Groups
The solubility of a molecule in water is often influenced by the presence of polar functional groups. These groups contain atoms with significantly different electronegativities, such as oxygen, nitrogen, or halogens, which create a dipole. This means one part of the molecule becomes slightly positive, while the other part becomes slightly negative. This separation of charge is crucial in determining how molecules interact with water, a polar solvent.

When comparing the solubility of two substances, it's important to identify which functional groups they possess. Polar groups, like hydroxyl (OH), carboxyl (COOH), and nitrile (CN), enhance water solubility since they can engage in interactions with water molecules.
  • The polarity of these functional groups allows them to dissolve better in water, breaking into smaller molecules that interact readily with water.
  • Molecules lacking polar functional groups, like hydrocarbons, tend to be less soluble in water, as they do not have the dipole-dipole interactions needed to interact with water molecules.
Hydrogen Bonding
Hydrogen bonding is a critical type of dipole-dipole interaction occurring between molecules. It involves a hydrogen atom, which is covalently bonded to a highly electronegative atom like oxygen, nitrogen, or fluorine, forming a partial positive charge. This slightly positive hydrogen then attracts a nearby electronegative atom in a different molecule.

Water molecules are excellent at forming hydrogen bonds due to the presence of highly electronegative oxygen atoms bonded to hydrogens. Therefore, substances capable of hydrogen bonding tend to dissolve well in water.
  • Molecules with functional groups like alcohols (OH) and amines (NH) can engage in hydrogen bonding, significantly enhancing their solubility.
  • Comparing molecules' ability to hydrogen bond gives insight into their relative water solubility. Stronger hydrogen bonding corresponds to better solubility in water.
Organic Chemistry
Organic chemistry is the study of carbon-containing compounds and it's crucial for understanding solubility, especially for organic compounds in water. The interplay between the molecule's structure, functional groups, and existing bonds like hydrogen bonds dictate how well a molecule can dissolve in different solvents.

In organic chemistry, identifying and understanding these interactions provides insights into predicting a compound's behavior when mixed with solvents like water. Recognizing these features in molecules helps in the synthesis of new compounds and in enhancing the properties of existing ones.
  • Most organic compounds have carbon backbones, and the solubility comes down to what other elements and functional groups are attached to this backbone.
  • The presence and specific arrangement of polar functional groups can either hinder or promote the interaction with water, affecting solubility.
Molecular Polarity
Molecular polarity plays a central role in determining a compound's solubility in water. A molecule is polar if it has a net dipole moment, with one part carrying a slight excess of positive charge and the other part carrying a slight excess of negative charge. This generally arises from an uneven distribution of electrons across the molecule, usually due to polar covalent bonds.

Water is a polar molecule itself, enabling it to dissolve other polar substances effectively. "Like dissolves like" is a useful rule of thumb for understanding solubility. Polar molecules mix well with water, while nonpolar molecules do not.
  • In polar molecules, the attraction between partial charges in water and the solute leads to dissolution.
  • The overall structure, including the orientation and number of polar bonds, impacts how molecules interact with water.
Understanding these principles helps in predicting which substances are likely to be soluble in water and which are not.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Rationalize the trend in water solubility for the following simple alcohols:

An aqueous solution containing glucose has a vapor pressure of 19.6 torr at \(25^{\circ} \mathrm{C}\). What would be the vapor pressure of this solution at \(45^{\circ} \mathrm{C}\) ? The vapor pressure of pure water is 23.8 torr at \(25^{\circ} \mathrm{C}\) and 71.9 torr at \(45^{\circ} \mathrm{C} .\) If the glucose in the solution were substituted with an equivalent amount (moles) of \(\mathrm{NaCl}\), what would be the vapor pressure at \(45^{\circ} \mathrm{C}\) ?

You make \(20.0 \mathrm{~g}\) of a sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) and \(\mathrm{NaCl}\) mixture and dissolve it in \(1.00 \mathrm{~kg}\) water. The freezing point of this solution is found to be \(-0.426^{\circ} \mathrm{C}\). Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

The vapor pressure of a solution containing 53.6 \(\mathrm{g}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) in 133.7 \(\mathrm{g}\) ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 113 torr at \(40^{\circ} \mathrm{C}\) . Calculate the vapor pressure of pure ethanol at \(40^{\circ} \mathrm{C}\) assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.

A solution is prepared by mixing 0.0300 mole of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and 0.0500 mole of \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) at \(25^{\circ} \mathrm{C}\) . Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions at \(25^{\circ} \mathrm{C}\) . At \(25^{\circ} \mathrm{C}\) , the vapor pressures of pure \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and pure \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) are 133 and 11.4 torr, respectively.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free